Abstract Algebra, Midterm II Solutions

1.) First note that in disjoint cycle form b = (1342). This 4-cycle has orderfour. Also note that b42 = b40b2 = (b4)10b2 = e10b2 = b2. One computesthat b2 = (14)(32); thus b42 = b2 = (14)(32).

2.) Let us assume that there exists an onto homomorphism φ from Z8 ⊕Z3

onto Z4 ⊕ Z6. Note that we have |Z8 ⊕ Z3| = 24 and |Z4 ⊕ Z6| = 24. Sincethese groups have the same number of elements and φ is onto, φ must be one-one. We assumed that φ is a homomorphism. So it is operation-preserving.This means that φ is an isomorphism. By Theorem 6.2(5), we know that|φ(g)| = |g| for every g ∈ Z8 ⊕ Z3. Let us consider the order of the element(1, 1) ∈ Z8⊕Z3. We have that |(1, 1)| has order lcm(8, 3) = 24. But note thatif (g1, g2) is an element of Z4 ⊕ Z6, we have |(g1, g2)| = lcm(|g1|, |g2|) ≤ 12because |g1| can only be 1, 2 or 4 and |g2| can only be 1,2, 3 or 6. In otherwords, Z4⊕Z6 doesn’t has an element of order 24. On the contrary, Z8⊕Z3

has an element of order 24. This contradicts Theorem 6.2(5). Hence thereis no isomorphism between Z8 ⊕ Z3 and Z4 ⊕ Z6.

3.) This is Chapter 7, number 26. We use Corollary 2 of the Lagrange’sTheorem to answer this question. Let g be a nonidentity element of |G|.Since G is finite, |g| divides |G|. In other words, we get that |g| divides 8.We have the following possibilities: |g| = 1, |g| = 2, |g| = 4 or |g| = 8.First, the identity element, e, is the only element with order 1. So there areseven other elements to consider. If |g| is 2, we are done. If |g| is 4, thenwe consider the element g2 (which is not e since |g| 6= 2). Note that wehave (g2)2 = e. So, we get |g2| = 2. If |g| is 8, then we consider g4 (again,g4 6= e). Since we calculate that (g4)2 = e, we find that |g4| = 2. In anycase, we obtain an element of order 2. Therefore , if |G| is 8, then G has anelement of order 2.

4.) (a) Since (23)2 = e, |H| = 2. Since |S4| = 24 and A4 has half theelements of S4, |A4| = 12. Thus the number of cosets of H in A4 is 12/2 = 6.

(b) Since H = {e, (23)}, we see that (12)(34)H = {(12)(34), (12)(34)(23)}.Putting the second element in disjoint cycle form gives (12)(34)H = {(12)(34), (2431)}.

(c) Consider (12)(34)H[(12)(34)]−1. First [(12)(34)]−1 = (34)−1(12)−1 =(34)(12) = (12)(34). So (12)(34)H(12)(34) = {(12)(34)(12)(34), (2431)(12)(34)}.Calculating these products gives (12)(34)H(12)(34) = {e, (14)}. This showsthat H is not normal in A4 since xHx−1 is not equal to H for x = (12)(34)which is an element of A4.

1

5.) We need to find the smallest k such that 5k mod 12 is in 〈4〉 = {0, 4, 8}.The smallest such number is k = 4, since then 20 mod 12 = 8. (Note 10mod 12 = 2 and 15 mod 12 = 3.) So the order of 5 + 〈4〉 is 4.

6.) Since φ(7) = 17 we know that 7φ(1) = 17, or φ(1) = (7−1 · 17) mod 20.In Z20 the multiplicative inverse of 7 is 3 since 3 · 7 mod 20 = 1. Thusφ(1) = (3 · 17) mod 20 = 51 mod 20 = 11. So φ(x) = 11 · x mod 20. Alsocheck that φ(7) = (11 · 7) mod 20 = 77 mod 20 = 17.

2